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# Further results on common zeros of the solutions of two differential equations

- Asim Asiri
^{1}Email author

**2012**:222

https://doi.org/10.1186/1029-242X-2012-222

© Asiri; licensee Springer 2012

**Received:**8 May 2012**Accepted:**17 September 2012**Published:**4 October 2012

## Abstract

### Purpose

Two problems are discussed in this paper. In the first problem, we consider one homogeneous and one non-homogeneous differential equations and study when the solutions of these differential equations can have (nearly) the same zeros. In the second problem, we consider two linear second-order differential equations and investigate when the solutions of these differential equations can take the value 0 and a non-zero value at (nearly) the same points.

### Method

We apply the Nevanlinna theory and properties of entire solutions of linear differential equations.

### Conclusion

In the first problem, the results determine all pairs of such equations having solutions with the same zeros or nearly the same zeros. Regarding the second problem, the results also show all pairs of such equations having solutions taking the value 0 and a non-zero value at (nearly) the same points.

## Keywords

- Nevanlinna theory
- differential equations

## 1 Introduction

There has been much research [1–8] on zeros of solutions of linear differential equations with entire coefficients. The principal paper [9] that was published in 1982 by Bank and Laine has stimulated many studies on this kind of problems. The reader is referred to [10–12] for background on some applications of the Nevanlinna theory. We use the standard notation of the Nevanlinna theory from [13].

In 1955, Wittich [12] proved the following theorem.

**Theorem 1.1** *If* *f* *is a non*-*trivial solution of* ${w}^{\u2033}+Aw=0$, *i*.*e*., $f\not\equiv 0$ *and* $A\not\equiv 0$ *is entire*, *then we have*:

(i) $T(r,A)=S(r,f)$.

(ii) *If* *f* *has finite order*, *then* *A* *is a polynomial*.

*If*

*a*

*is a non*-

*zero complex number*,

*then*

*f*

*takes the value*

*a*

*infinitely often*,

*and in fact*,

*outside a set of finite measure*,

*Let*

*P*

*be a polynomial of degree*

*n*,

*and let*

*w*

*be a non*-

*trivial solution of the equation*(1).

*Then*,

*w*

*has order of growth equal to*$\frac{n+2}{2}$.

*Moreover*,

*if*

*w*

*is a solution of*(1)

*which has infinitely many zeros*,

*then we have*

Our previous paper [14] studied homogeneous linear differential equations having solutions with nearly the same zeros and proved several results, including the following.

**Theorem 1.3** [14]

*Let*$P\not\equiv 0$

*be a polynomial of degree*

*n*.

*Let*$w\not\equiv 0$

*be a solution of*(1).

*Assume that*

*w*

*has infinitely many zeros*.

*Suppose that we have a solution*$v\not\equiv 0$

*of the differential equation*

*such that*

*A*

*is an entire function and*$N(r)$

*counts zeros of*

*v*

*which are not zeros of*

*w*

*and zeros of*

*w*

*which are not zeros of*

*v*.

*Assume that*

*Then* $\frac{v}{w}$ *is a constant and* $A=P$.

The paper [14] includes further results for homogeneous linear differential equations, and the corresponding problem where *P* is a transcendental entire function of finite order is studied in [15].

Recently, the same problem, but with non-homogeneous first-order differential equations, has been studied in [16], including the following result.

**Theorem 1.4** [16]

*Assume that*${v}^{\prime}=Av+B$

*and*${w}^{\prime}=Cw+D$,

*where*

*A*,

*B*,

*C*

*and*

*D*

*are entire functions of order less than*1

*and*

*v*,

*w*

*are transcendental functions*.

*Assume that*$v=Lw$,

*where*

*L*

*has finitely many zeros and poles*,

*and*

*Then the following conclusions hold*.

(I) *If* *L* *is a rational function*, *then* $A\equiv C$, *L* *is a constant and* $B=LD$.

(II) *If* *L* *is a transcendental function*, *then one of the following cases holds*: *If*, *in addition*, *L* *has finite order in* case (ii), *then* *A*, *B*, *C*, *D* *are polynomials and so is* ${A}_{1}$.

(i) $B\equiv D\equiv 0$ *and* *v*, *w* *have no zeros*.

*and*$B/A$, $D/C$

*are non*-

*zero constants*,

*and*

*where* ${c}_{j}\in \mathbb{C}$, ${A}_{1}^{\prime}=A$ *and* $L=(\mathit{\text{constant}}){e}^{{A}_{1}}$.

In this paper, our first result (Theorem 2.1) looks at the same problem but with one homogeneous and one non-homogeneous differential equations. In particular, we consider the first equation to be homogeneous of the second-order with a polynomial coefficient and the second equation to be non-homogeneous of the first-order with entire coefficients.

A further result (Theorem 2.2) studies the case where the solutions of two second-order homogeneous differential equations can take the value 0 and a non-zero value at (nearly) the same points.

## 2 Our results

Our first result is the following theorem.

**Theorem 2.1**

*Suppose that*$P\not\equiv 0$

*is a polynomial of degree*

*n*,

*and*

*w*

*solves*(1),

*and*$w\not\equiv 0$

*has infinitely many zeros*.

*Suppose that*$v\not\equiv 0$

*solves*

*where*

*A*,

*B*

*are entire and*$AB\not\equiv 0$,

*and*

*Suppose that* $L=\frac{w}{v}$ *has finitely many zeros and poles* (*i*.*e*., *w* *and* *v* *have the same zeros with finitely many exceptions*).

*Then*

*A*

*is a polynomial and there exists a polynomial*

*Q*

*such that*

*and*

*where* ${A}_{1}^{\prime}=A$ *and* ${c}_{1}$, ${c}_{2}$ *are constants*.

**Example 2.1**Take

*Q*to be a polynomial. Let

So, we have $P=-({Q}^{\u2033}+{Q}^{\mathrm{\prime}2})$.

*v*has the same zeros as

*w*. Now, we have

where $A={A}_{1}^{\prime}$.

For example, let ${A}_{1}=2Q$.

We now state our second result.

**Theorem 2.2** *Suppose that* $P\not\equiv 0$ *is a polynomial of degree* *n*, *and* *A* *is an entire function*, *and suppose that* *w* *solves* (1) *and* *v* *solves* (3), *and* $vw\not\equiv 0$. *Let* $v-1$ *and* *w* *have*, *with finitely many exceptions*, *the same zeros and the same multiplicities*. *Then one of the following holds*.

(A) *w* *has finitely many zeros and* *v* *is a polynomial and* $A=0$.

*w*

*has infinitely many zeros and*

*P*,

*A*

*are non*-

*zero constants and*$\frac{v-1}{w}$

*is non*-

*constant and*

*where* *σ*, ${\lambda}_{1}$, ${\lambda}_{2}$, ${\lambda}_{3}$ *are non*-*zero constants*.

**Example 2.2**If $w={e}^{z}-{e}^{-z}$ and $v={e}^{2z}$, then

Hence, $v-1$ has the same zeros as *w*. Here $P=-1$ and $A=-4$.

**Example 2.3**We give an example to show that the zeros of $v-1$ and

*w*must necessarily have the same multiplicities. To show this, let

Then $w=0\iff \frac{z}{2}=k\pi $, where $k\in \mathbb{Z}$.

Also $v=1\iff z=k2\pi $, where $k\in \mathbb{Z}$.

Therefore, *w* and $v-1$ have the same zeros but the zeros are simple for *w*, double for $v-1$. Here, $P=\frac{1}{4}$ and $A=1$.

## 3 Proof of Theorem 2.1

*Proof*We have

*L*, and by using (12) and (13), we get

The next step is to estimate the growth of *M*.

**Claim 3.1** *We claim that* $T(r,M)=o({r}^{(n+2)/2})$.

To show this, we know that $N(r,m)=O(logr)$ since *M* has finitely many poles.

where ${A}_{1}^{\prime}=A$.

*c*such that

We also have $N(r,M)=O(logr)$.

This completes the proof of Claim 3.1.

because otherwise we can write $v=-{M}_{2}/{M}_{1}$ to get a contradiction.

*B*to get

*B*has finitely many zeros. Then we can write

*B*in the form

where ${P}_{1}$, ${P}_{2}$ are polynomials.

where ${R}_{1}$ is rational.

where ${R}_{2}$ is rational and ${R}_{2}=-\frac{1}{2}{R}_{1}$.

Thus, $\rho (H)<\mathrm{\infty}$, *H* is entire, and *B* has no zeros.

Therefore, *A* is a polynomial.

*B*has no zeros, from (20) we can write

where *Q* is a polynomial.

*w*and

*H*solve the same equation and are linearly independent (because

*w*has zeros but

*H*does not), we can write

where ${c}_{1}$ is a constant and *Q* is a polynomial.

where ${c}_{2}$ is a constant and ${A}_{1}^{\prime}=A$.

Now, from (23) and (24), we notice that *w* and *v* have the same zeros.

This completes the proof of Theorem 2.1. □

## 4 A lemma needed to prove Theorem 2.2

In order to prove Theorem 2.2, we must state and prove the following lemma. We include a proof for completeness.

**Lemma 4.1**

*Let*${P}_{1},\dots ,{P}_{n}\in \mathbb{C}$

*be distinct*,

*and let*${A}_{1},\dots ,{A}_{n}$

*be rational functions such that*

*Then there exists* $k\in \{1,\dots ,n\}$ *such that* ${P}_{k}=0$ *and* ${A}_{k}=1$, *and* ${A}_{j}=0$ *for* $j\ne k$.

*Proof* The proof is by induction. It is obvious that the lemma is true when $n=1$.

Now, we have two cases to consider.

*k*such that ${B}_{k}\not\equiv 0$. Without loss of generality, let $k=1$, then we can write

Since we assumed the lemma is true for $m\le n-1$, there exists $j\in \{2,\dots ,n\}$ such that ${P}_{j}-{P}_{1}=0$. But this contradicts our assumption that ${P}_{1},\dots ,{P}_{n}$ are distinct.

*j*,

*i.e.*,

But this contradicts the fact that ${P}_{j}\ne 0$.

*k*)

and ${P}_{k}=0$ and ${A}_{k}=1$. □

## 5 Proof of Theorem 2.2

In particular, if *w* has finitely many zeros, then *v* is a polynomial, which gives $A=0$. This completes the proof of part (A) in the conclusion.

Assume henceforth that *w* has infinitely many zeros. Then (26) implies that $\rho (v)\le (n+2)/2$, and so *A* is a polynomial of degree at most *n* by the Wiman-Valiron theory [17]. Also, $A\not\equiv 0$ since $v-1$ has infinitely many zeros.

Now, two cases have to be considered.

*P*is a non-zero constant; then $n=0$ and

*A*is constant. Therefore, we can write

where $\alpha ,\beta \in \mathbb{C}\setminus \{0\}$, ${c}_{j},{d}_{j}\in \mathbb{C}$ and ${c}_{j}\ne 0$ ($j=1,2$).

*w*and $v-1$ have the same zeros with finitely many exceptions, we can write

where *δ* is constant.

and $\alpha +\gamma $, $-\alpha +\gamma $, *β*, −*β*, 0 cannot all be different.

We must now try six cases:

I(a): If $\alpha +\gamma =\beta $ and $-\alpha +\gamma =-\beta $, then $\gamma =0$ and $\alpha =\beta $. But this contradicts (30). Thus, this case cannot happen.

where ${h}_{1}$, ${h}_{2}$ are constants, which yields ${d}_{2}=0$. Putting this in (27) gives (7) with $\sigma =\gamma $.

There are four more cases:

I(c): $\alpha +\gamma =-\beta $ and $-\alpha +\gamma =\beta $.

I(d): $\alpha +\gamma =-\beta $ and $-\alpha +\gamma =0$.

I(e): $\alpha +\gamma =0$ and $-\alpha +\gamma =\beta $.

I(f): $\alpha +\gamma =0$ and $-\alpha +\gamma =-\beta $.

It is easy to check that case I(c) is impossible and cases I(d), I(e), I(f) all lead to (7) with $\sigma =\gamma $.

From these cases, we find that $\gamma \ne 0$, and so $\frac{v-1}{w}={e}^{\gamma z+\delta}$ is non-constant. Also, we have (7) and case (B) of the conclusion.

*P*is non-constant. We will show that this leads to a contradiction. Let

where *L* is a rational function and *Q* is an entire function.

From (26), we have $\rho (v)<\mathrm{\infty}$, and so *Q* is a polynomial.

Now, we have two cases to consider.

*M*is constant, then either $A=P$ and $A=0$, so that $P=0$, which is a contradiction, or

is a rational function, which is a contradiction since *w* has infinitely many zeros.

*M*is non-constant, then ${M}^{\prime}\not\equiv 0$. Therefore,

where $\frac{{M}^{\u2033}}{2{M}^{\prime}}+\frac{(A-P)M}{2{M}^{\prime}}$ is rational because $\frac{{M}^{\prime}}{M}=\frac{{L}^{\prime}}{L}+{Q}^{\prime}$ is rational and $\frac{{M}^{\u2033}}{M}$ is rational, and so $\frac{{M}^{\u2033}}{{M}^{\prime}}=\frac{{M}^{\u2033}}{M}/\frac{{M}^{\prime}}{M}$ is rational.

are rational functions and *Q* is a polynomial.

Now, we have two cases to consider:

which is a contradiction since *w* has infinitely many zeros.

*w*has finitely many zeros, a contradiction. Also,

where $T=1/S$ is a rational function.

So, *G* solves (1), and since $P\not\equiv 0$ and is a polynomial of degree *n*, we see that *G* is a transcendental entire function with finitely many zeros and has order $(n+2)/2$.

*w*and

*G*solve the same equation but

*w*has infinitely many zeros and

*G*has finitely many zeros,

*w*and

*G*are linearly independent, and we can write

*c*is a non-zero constant. So,

where $H=L/T$ is a rational function.

Now, we can assume that $c=1$ because if $c\ne 1$, we can multiply *w* by $1/c$.

is a rational function.

where ${U}_{1}={K}^{\prime}+{K}^{2}+A$ and ${U}_{2}={H}^{\prime}-{K}^{\prime}+KH-{K}^{2}$.

*v*is transcendental and ${U}_{1}$, ${U}_{2}$ are rational functions, we must have

**Claim 5.1** *We claim that* $H\equiv K$.

To show this, let $H\not\equiv K$.

where *a* is a constant. But this contradicts the fact that *H* and *K* are rational functions and *G* is a transcendental function. This completes the proof of Claim 5.1.

*v*has finitely many zeros, so we can write

where ${P}_{1}$, ${Q}_{1}$ are polynomials, ${P}_{1}\not\equiv 0$, and ${Q}_{1}$ is non-constant because *v* is transcendental.

where ${R}_{1}={P}_{1}/L\not\equiv 0$, ${R}_{2}=-1/L\not\equiv 0$ are rational functions and ${S}_{1}={Q}_{1}-Q$, ${S}_{2}=-Q$ are polynomials.

Therefore, ${J}_{1}={J}_{2}=0$ because otherwise ${e}^{{Q}_{1}}=-{J}_{2}/{J}_{1}$ or ${e}^{-{Q}_{1}}={J}_{1}/{J}_{2}$. Thus, ${R}_{1}{e}^{{S}_{1}}$, ${R}_{2}{e}^{{S}_{2}}$ both solve ${y}^{\u2033}+Py=0$ and have finitely many zeros, and they are linearly independent.

Hence, *P* is constant by [9], which contradicts our assumption in Case (II) that *P* is non-constant. □

## Declarations

### Acknowledgements

The author would like to thank his supervisor Prof. Jim Langley for his support and guidance. Also, he would like to thank King Abdulaziz University for financial support for his PhD study.

## Authors’ Affiliations

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